2021.2.23--vj补题

B - B

 CodeForces - 699B 

题目：

You are given a description of a depot. It is a rectangular checkered field of n × m size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*").

You have one bomb. If you lay the bomb at the cell (x, y), then after triggering it will wipe out all walls in the row x and all walls in the column y.

You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall.

Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 1000) — the number of rows and columns in the depot field.

The next n lines contain m symbols "." and "*" each — the description of the field. j-th symbol in i-th of them stands for cell (i, j). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is occupied by a wall.

Output
If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes).

Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them.

Examples
Input
3 4
.*..
....
.*..
Output
YES
1 2
Input
3 3
..*
.*.
*..
Output
NO
Input
6 5
..*..
..*..
*****
..*..
..*..
..*..
Output
YES
3 3

题意：输入一个n*m的矩阵，由*和. 分别表示墙和路，设置一个炸弹可以炸掉所在的行和列，问能不能放置一枚炸弹，炸掉所有的墙，若能，则输出“”YES“”，并在下一行输出炸弹坐标，否则输出“NO”

思路：用二维数组s[][] 输入，sum记录下墙的总数，同时用数组x[ ], y[ ] 记录下每一行和每一列的墙的总数目，最后遍历时对比

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,m;
    cin>>n>>m;
    int x[1010]={0},y[1010]={0},sum=0;
    char s[1010][1010];
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            cin>>s[i][j];
            if(s[i][j]=='*')
            {
                sum++;
                x[i]++;
                y[j]++;
            }
        }
    }
    int q=0;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if((s[i][j]=='*'&&x[i]+y[j]-1==sum)||(s[i][j]!='*'&&x[i]+y[j]==sum))
            {
                q=1;
                cout<<"YES"<<endl;
                cout<<i+1<<" "<<j+1<<endl;
                break;
            }
        }
        if(q==1)break;
    }
    if(q==0)cout<<"NO"<<endl;
}